# Understanding the proof of: if A Connected and $A \subset B \subset \bar{A} \Rightarrow B$ is connected

I shall write the proof of this following theorem, which i have found on the internet. i shall mark the part which i do not understand.

The proof is done by contradiction:

Let $$B$$ be not connected, thus there exists open, disjunct none empty sets $$U, V$$ such that $$B \subset U \cup V$$ with $$B \cap U$$ and $$B \cap V$$ non-empty.
Because $$A \subset B$$ then obviously $$A \subset U \cup V$$.
Furthermore are the intersections $$A \cap V$$ and $$A \cap U$$ none empty. For if for example $$A \cap U = \emptyset$$ then $$A \subset X-U$$ then because $$X - U$$ is closed (Because $$U$$ is open) we have

1. $$B \subset \bar{A} \subset X-U$$

This is however is a contradiction to the fact that $$B \cap U \neq \emptyset$$

Analog we find that $$A \cap V$$ is not empty

But then $$V , U$$ would disassemble A then A would not be connected which is a contradiction.

What i do not understand is marked as equation (1). Why does $$\bar{A} \subset X-U$$ ? The rest of the proof is clear to me. Thanks for your help.

The closure $$\overline{A}$$ of $$A$$ in $$X$$ is the smallest closed subset of $$X$$ containing $$A$$. More precisely, it is the intersection of all closed subsets of $$X$$ containing $$A$$ and thus contained in every closed subset containing $$A$$. In your case the set $$X - U$$ is closed so that $$A \subset X - U$$ implies $$\overline{A} \subset X - U$$.
• Interesting, does Generally speaking $A \subset B$ where as B closed thus we write $A \subset B = \bar {B}$ imply that $\bar{A} \subset B = \bar {B}$ ? May 29 at 10:40
• @MadSpaces : Maybe even a bit more general, $A \subset B$ implies $\overline{A} \subset \overline{B}$ for any $A$ and $B$. May 29 at 10:44